A ray of light travels from Liquid helium (n=1.03) into a second material. You gradually increase the angle of incidence until, at 83.6°, the light no longer passes into the second material. What is the index of refraction of the second material?
sinTheta=n/1.03
n/1.03=.994
n= 1.028
http://www.physicsclassroom.com/class/refrn/Lesson-3/The-Critical-Angle
To find the index of refraction of the second material, we can use Snell's Law, which relates the angle of incidence and the angle of refraction to the indices of refraction of the two materials. Snell's Law can be stated as:
n1 * sin(i) = n2 * sin(r)
Where:
n1 is the index of refraction of the first material (liquid helium)
n2 is the index of refraction of the second material
i is the angle of incidence
r is the angle of refraction
In this case, we know the index of refraction of the first material (liquid helium) is 1.03. We also know that at an angle of incidence of 83.6°, the light no longer passes into the second material. At this point, the angle of refraction is 90° since light no longer enters the second material.
We can rewrite Snell's Law to solve for n2:
n2 = (n1 * sin(i)) / sin(r)
Plugging in the known values, we get:
n2 = (1.03 * sin(83.6°)) / sin(90°)
To evaluate this further, we need to convert the angles from degrees to radians. The formula to convert degrees to radians is:
Radian = Degree * (π/180)
Substituting the values, we get:
n2 = (1.03 * sin(83.6 * π/180)) / sin(90 * π/180)
Evaluating this expression, we find that the index of refraction of the second material is approximately 1.48.